Let $f$ be a vector-valued function defined by $f(t)=(\log_3(t),t^4+2t)$. Find $f'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{3t}+4t+2$ (Choice B) B $\left(\dfrac{1}{3t},4t+2\right)$ (Choice C) C $\left(\dfrac{1}{t\ln(3)},4t^3+2\right)$ (Choice D) D $\left(-\log_3(t),4t^3+2\right)$
Answer: $f$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $f(t)=(\log_3(t),t^4+2t)$. Let's differentiate the first expression: $\dfrac{d}{dt}[\log_3(t)]=\dfrac{1}{t\ln(3)}$ Let's differentiate the second expression: $\dfrac{d}{dt}(t^4+2t)=4t^3+2$ Now let's put everything together: $\begin{aligned} f'(t)&=\left(\dfrac{d}{dt}[\log_3(t)],\dfrac{d}{dt}(t^4+2t)\right) \\\\ &=\left(\dfrac{1}{t\ln(3)},4t^3+2\right) \end{aligned}$ In conclusion, $f'(t)=\left(\dfrac{1}{t\ln(3)},4t^3+2\right)$.